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Inductor vs Capacitor energy transfer prediction

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Armagdn03

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So, If I have an inductor with a steady current through it, and its power is cut, and the collapse fills a capacitor, (picture a boost converter) can you calculate the voltage that the capacitor will fill to?

I was thinking, the energy stored in a coil is P=.5L x I^2 where L is inductance, and I is current,

And since the equation for energy stored in a cap is P=.5C x I^2 so...

set one equal to the other and you can find how much current a capacitor discharged into a coil will create, and how much voltage a coil discharging into a cap will create, (minus losses of course).

Seem like a reasonable plan of attack?
 

dknguyen

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The capacitor does not stay charged. As the capacitor charges up, it's voltage increase will slow reaching DC when the capacitor is done charging. This slowing voltage increase approaching DC translates to a deceasing frequency. Remember that inductors appear as lower impedance for lower frequencies (becoming a short-circuit at DC). This low impedance across the capacitor will drain energy from the capacitor, so the energy will flow from the capacitor back into the inductor "magnetizing" it. The energy jumps back and forth between L and C producing an oscillating LC resonant circuit.

That's one reason boost converters have diodes- to keep that energy in the capacitor.
 
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Armagdn03

New Member
Right I am quite familiar with ressonant circuits, the need for diodes when restricting flow, standing wave voltage ratios and the whole nine yards of why they all do what they do, except, I would like to break down the process into steps, being able to describe the energy from one to the next, this is why I broke the problem down into joules in order to figure it out. I have never seen anybody else try this problem so thought I would ask.

This is useful for delay lines, transmission line models, filter design, boost converters, buck converters and many other things!
 

dknguyen

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Well, without losses the oscillation of energy between the L and C is supposed to continue forever. So I think L energy = C energy is fair.
 
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So, If I have an inductor with a steady current through it, and its power is cut, and the collapse fills a capacitor, (picture a boost converter) can you calculate the voltage that the capacitor will fill to?
I wouldn't do it that way. Set up the La Place transform of the circuit, and find the time domain solution. That'll give you your answer.
 

dknguyen

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Are you just looking for a description of the waveforms in an LC circuit? What you have to realize is that analyzing circuits with inductors and capacitors in them is no different than analyzing a circuit with resistors. It's done exactly the same way- the equations for the voltage and current for inductors and capacitors is just more complicated than resistors is all, and make it more involved to solve the equation.

Like Miles says, it is a loop analysis of an LC circuit.

In the same way:
resistors are V=IR
inductors are V = L*di/dt
capacitor are I = C*dv/dt -> dV = (I/C)dt -> V = ∫(I/C)dt
If you just have an L and a C connected in a loop, Kirchoff's voltage law around the loop works out to be:

0 = Vinductor - Vcapacitor
0 = L*dI/dt - ∫(I/C)dt


This is a differential equation, but is not in the form we are used to solving (for the form required by most methods to solve the equation) because it has an integral in it. The form we are used to solving has all derivatives in it. So to change this equation into that form we take the derivative of both sides.
0 = L*d^2I/dt^2 - I/C

Now it has all derivatives and is almost in the form we want. We can solve it in this form now, but usually we like to remove the coefficient from the term with the highest derivative (some methods to solve require it be like this, others do not). We divide everything by L to remove the L from the term of the highest order (the term with the second derivative):
0 = d^2I/dt^2 - I/LC

This differential equation is now in a form that matches the form required by most methods used to solve it. This is a differential equation where the variables are I and and it's derivative, dI/dt. You need to solve it using partial differential equation methods (Laplace transforms is one method to solve this equation). You simulate the disconnection of a power source by setting the initial conditions in your solution to be that where the capacitor is charged up to a certain voltage or where the inductor has an initial current flowing through it.

THe solution you get for I and dI/dt will be an equation that describes the time behaviour current in the circuit. Plugging it back into the Capacitor and Inductor voltage formulas will give you their voltage behaviour over time. Because it is an oscillator your solution will be some kind of periodic waveform (for both current and the voltages of the capacitor and inductor).

An example of this math that is carried out to completion is under "Time Domain Solution" is one example of how to approach the problem. This particular method does not use Laplace transforms to solve it. They leave out a lot of details on what goes on as they solve the differential equation, but a lot of people prefer Laplace transforms instead. If you do not understand what they are doing you need to learn how to solve differential equations- that is the key.
http://en.wikipedia.org/wiki/LC_circuit
 
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crutschow

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And since the equation for energy stored in a cap is P=.5C x I^2 so...
It's may be just a typo but the energy stored in a cap is P=.5C x V².

And yes, for a lossless system you can just equate the inductor energy to the capacitor energy to get the peak current in the inductor and the peak voltage on the capacitor as the circuit oscillates. You don't need any more complex math then that.
 

MrAl

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Well, without losses the oscillation of energy between the L and C is supposed to continue forever. So I think L energy = C energy is fair.

Hello there,


I have to agree that the oscillation continues indefinitely without any
losses in the circuit, and those losses absent would have to also
be any electromagnetic radiation, so that would be very hard to
accomplish but that's pure theory for ya.
 

Sceadwian

Banned
The real world doesn't work like that though, even in the most ideal of circuits there is always loss. It's best to keep this in mind from the start even if you're learning theory. As the first time you put a circuit into practice you'll get results different from pure theory.
 

MrAl

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Most Helpful Member
Who was it that said,

"In theory theory works every time, in practice it doesnt".
 
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